If value of $M = 30$, then for which of the following series value of $M$ not satisfied properly? I. $18, M, 50, 97, 208, 444$ II. $-50, -49, -45, -35, -16, \frac{M}{2}$ III. $8, 5, 6.5, M, 110, 1319$

### Concept & Number Series Anomalies Test the given value in each sequence by checking successive differences (first and second order) to see if a consistent mathematical pattern breaks. ### Step-by-Step Solution * **Given:** We need to test if $M = 30$ fails in any of the three series. * **Testing Series I:** $18, 30, 50, 97, 208, 444$. * First differences: $30-18=12$; $50-30=20$; $97-50=47$; $208-97=111$; $444-208=236$. * Second differences: $20-12=8$; $47-20=27$; $111-47=64$; $236-111=125$. * These are cubes: $2^3, 3^3, 4^3, 5^3$. The pattern holds perfectly. * **Testing Series II:** $-50, -49, -45, -35, -16, \frac{M}{2}$. * If $M=30$, the last term is $\frac{30}{2} = 15$. * First differences: $1, 4, 10, 19, 31$. * Second differences: $3, 6, 9, 12$. This is a perfect arithmetic progression. The pattern holds. * **Testing Series III:** $8, 5, 6.5, M, 110, 1319$. * If $M=30$, let's check the transitions. * $8 \times 0.5 + 1 = 5$ * $5 \times 1 + 1.5 = 6.5$ (or $5 \times 1.5 - 1 = 6.5$) * There is no continuous multiplication/addition sequence using $30$, $110$, and $1319$ that fits standard series logic (e.g., $110 \times 12 - 1 = 1319$ demands a much steeper curve than $M=30$ provides). * Since I and II perfectly accommodate $M=30$, III must be the one that does not. ### Exam Strategy & Shortcut In "Which is NOT satisfied" questions, start with the easiest series to verify. Series II is a simple double-difference sequence. Verifying I and II quickly leaves III as the only possible answer by elimination. ### Common Pitfall Wasting time trying to find the *exact* correct value for M in Series III. Once you prove I and II work perfectly, you can confidently choose "Only III". ### Final Answer Therefore, the correct answer is **Only III**.